A satellite is moving with a constant speed v in a circular orbit...

A satellite is moving with a constant speed

v

in a circular orbit about the earth. An object of mass

m

is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

Options

A $\frac{1}{2} m v^2$
B $m v^2$
C $\frac{1}{3} m v^2$
D $\frac{2}{3} m v^2$

Correct answer

B. $m v^2$

Step-by-step solution

Escape speed,

v_e=\sqrt{2} \times

orbital speed

=\sqrt{2} v

\therefore \quad

Kinetic energy of the object at the time of ejection

=\frac{1}{2} m v_e^2=\frac{1}{2} m(\sqrt{2} v)^2=m v^2

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